If two circles `(x+4)^(2)+y^(2)=1 and (x-4)^(2)+y^(2)=9` are touched extermally by a circle, then locus of centre of variable circle is

Options

A. (a) `(x^(2))/(15)-(y^(2))/(1)=1`
B. (b) `(x^(2))/(4)-(y^(2))/(12)=1`
C. (c) `(x^(2))/(1)-(y^(2))/(15)=1`
D. (d) `(x^(2))/(12)-(y^(2))/(4)=1`

Explanation

To solve the problem, we need to find the locus of the center of a variable circle that touches the given two circles externally. Let's break down the solution step by step.

### Step 1: Identify the centers and radii of the given circles
The equations of the circles are:
1. \((x + 4)^2 + y^2 = 1\)
2. \((x - 4)^2 + y^2 = 9\)

From these equations, we can identify:
- Circle 1: Center \(C_1 = (-4, 0)\), Radius \(r_1 = 1\)
- Circle 2: Center \(C_2 = (4, 0)\), Radius \(r_2 = 3\)

### Step 2: Set up the condition for external tangency
Let the center of the variable circle be \(C(h, k)\) and its radius be \(r\). For the variable circle to touch the first circle externally, the distance from \(C\) to \(C_1\) must equal the sum of their radii:
\[ \sqrt{(h + 4)^2 + k^2} = r + 1 \]

For the variable circle to touch the second circle externally, the distance from \(C\) to \(C_2\) must equal the sum of their radii:
\[ \sqrt{(h - 4)^2 + k^2} = r + 3 \]

### Step 3: Square both equations to eliminate the square root
Squaring both sides of the first equation:
\[ (h + 4)^2 + k^2 = (r + 1)^2 \]
Expanding this gives:
\[ h^2 + 8h + 16 + k^2 = r^2 + 2r + 1 \quad \text{(1)} \]

Squaring both sides of the second equation:
\[ (h - 4)^2 + k^2 = (r + 3)^2 \]
Expanding this gives:
\[ h^2 - 8h + 16 + k^2 = r^2 + 6r + 9 \quad \text{(2)} \]

### Step 4: Subtract the two equations
Subtract equation (1) from equation (2):
\[ (h^2 - 8h + 16 + k^2) - (h^2 + 8h + 16 + k^2) = (r^2 + 6r + 9) - (r^2 + 2r + 1) \]
This simplifies to:
\[ -16h = 4r + 8 \]
Rearranging gives:
\[ h = -\frac{r + 2}{4} \quad \text{(3)} \]

### Step 5: Substitute \(r\) in terms of \(h\) into one of the original equations
From equation (1):
\[ h^2 + 8h + 16 + k^2 = r^2 + 2r + 1 \]
Substituting \(r = -4h - 2\) from (3):
\[ h^2 + 8h + 16 + k^2 = (-4h - 2)^2 + 2(-4h - 2) + 1 \]
Expanding and simplifying will yield a relation between \(h\) and \(k\).

### Step 6: Find the locus equation
After substituting and simplifying, we find that the locus of the center \(C(h, k)\) is a hyperbola. The final equation will be of the form:
\[ \frac{x^2}{1} - \frac{y^2}{15} = 1 \]
This indicates that the locus of the center of the variable circle is a hyperbola.

### Final Answer
The locus of the center of the variable circle is given by:
\[ \frac{x^2}{1} - \frac{y^2}{15} = 1 \]